Bonding: OCR A A-level Chemistry

Understand ionic, covalent and metallic bonding, molecular shapes, bond angles, polarity and intermolecular forces.

Created by Dave Miller, Chemistry teacher, tutor and current AQA examiner with 15+ years of teaching experience. Independent resources for AQA and OCR A.

Ionic bonding & formulae

11 practice questions

Ionic formulae and common ions

Learning goalBalance charges and use correct brackets and smallest ratios.

🔓 CRACK IT

An ionic formula must have no overall charge. Choose the smallest whole-number ratio of ions that makes the positive and negative charges balance.

For example, one 2+ ion needs two 1- ions. Use the charges to find the ratio, then write the subscripts in the formula.

A compound ion stays together as a group. If you need more than one of that ion, put brackets around its formula before adding the subscript. The table below gives common ions to recognise.

Ionic formula: The smallest whole-number ratio of ions giving zero overall charge.

Group / ion	Charge or formula
Group 1	1+
Group 2	2+
Group 16	2−
Group 17	1−
Nitrate	NO₃^-
Carbonate	CO₃^2-
Sulfate	SO₄^2-
Hydroxide	OH^-
Ammonium	NH₄⁺
Zinc	Zn²⁺
Silver	Ag⁺

Worked example

Calcium nitrate

Write the formula from Ca²⁺ and NO₃^-.

Worked answer

One Ca²⁺ ion has charge +2.
Two nitrate ions give charge −2.
Keep each nitrate ion intact and use brackets.

Ca(NO₃)₂

EXAM TRAPUse brackets only when more than one compound ion is required. Simplify Ca₂O₂ to CaO; never change the internal formula of a compound ion.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 1

Complete the table of OCR A ions. [7]

Ion name	Formula and charge
Nitrate	…
Carbonate	…
Sulfate	…
Hydroxide	…
Ammonium	…
Zinc	…
Silver	…

Total: 7 marks

Show answer & marking points

NO₃^- nitrate [1]; CO₃^2- carbonate [1]; SO₄^2- sulfate [1]; OH^- hydroxide [1]; NH₄⁺ ammonium [1]; Zn²⁺ zinc [1]; Ag⁺ silver [1]

Exam-style question 2

Write the formula of each compound. Every formula must be neutral and in the smallest whole-number ratio. [6]

Part	Compound	Formula
a	Aluminium oxide	…
b	Calcium nitrate	…
c	Ammonium sulfate	…
d	Magnesium hydroxide	…
e	Zinc carbonate	…
f	Silver nitrate	…

(g) State when brackets are required around a compound ion. [1]

Total: 7 marks

Show answer & marking points

Aluminium oxide Al₂O₃ [1]; calcium nitrate Ca(NO₃)₂ [1]; ammonium sulfate (NH₄)₂SO₄ [1].

Magnesium hydroxide Mg(OH)₂ [1]; zinc carbonate ZnCO₃ [1]; silver nitrate AgNO₃ [1].

Use brackets only when more than one of a compound ion is needed, so the subscript applies to the whole ion [1]

✓ DONE WHENEvery formula is neutral, simplified and uses brackets correctly.

Ionic bonding and electron origin

Learning goalSeparate ion formation from ionic attraction and construct electron-origin diagrams.

🔓 CRACK IT

There are two ideas to keep separate. Electron transfer makes the ions. The attraction between the resulting positive and negative ions is the ionic bonding.

A metal atom loses electrons and becomes a positive ion. A non-metal atom gains electrons and becomes a negative ion. In a giant ionic lattice, each ion attracts surrounding ions of the opposite charge.

In a dot-and-cross diagram, use dots and crosses consistently to show where the electrons came from. Show outer-shell electrons, put each ion in square brackets and write its charge outside. Check both electron totals and the ratio of ions.

Ionic bonding: Electrostatic attraction between positive and negative ions.

EXAM TRAPA transfer arrow represents a process, not the ionic bond itself.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Quick check 3

(a) State the precise OCR A definition of ionic bonding. [2 marks]

(b) Explain why electron transfer is not itself the ionic bond. [2 marks]

Total: 4 marks

Show answer & marking points

(a) Ionic bonding is the electrostatic attraction [1] between positive and negative ions [1]. [2]

(b) Electron transfer is the process that forms the ions [1]; the bond is the attraction between the ions after transfer [1]. [2]

Exam-style question 4

Draw a dot-and-cross diagram for magnesium chloride, MgCl₂. Show outer-shell electrons only, electron origin, square brackets, charges and complete outer shells. [5 marks]

Total: 5 marks

Show answer & marking points

One Mg²⁺ ion and two Cl^- ions, all bracketed and correctly charged [2].

Each chloride ion has eight outer-shell electrons [1], with one electron on each chloride shown as transferred from Mg [1]. Magnesium has lost two electrons in total [1]. [5]

Exam-style question 5

Sodium forms Na⁺ ions and oxygen forms O^2- ions.

(a) Deduce the formula of sodium oxide. [1 mark]

(b) Draw a complete electron-origin diagram for sodium oxide. Show all ions, brackets, charges and outer-shell electrons. [6 marks]

Total: 7 marks

Show answer & marking points

(a) Two Na⁺ ions are needed for one O^2- ion, so the formula is Na₂O [1]. [1]

(b) Two bracketed Na⁺ ions and one bracketed O^2- ion [2]; oxide has eight outer-shell electrons [1], including two electrons shown as transferred from sodium [2]; electron totals and charges are consistent with Na₂O [1]. [6]

Exam-style question 6

Main-group elements X and Y form X³⁺ and Y^2- ions with complete octets. X has three outer-shell electrons and Y has six before ion formation.

(a) Deduce the formula of the compound formed. [2 marks]

(b) State the total number of electrons transferred when one formula unit forms. [2 marks]

(c) Draw a complete electron-origin diagram for one formula unit. Show the smallest ion ratio, complete outer shells, brackets and charges. [6 marks]

Total: 10 marks

Show answer & marking points

(a) Charge balance requires 2 × (+3) and 3 × (−2), so the formula is X₂Y₃ [2]. [2]

(b) Two X atoms each lose three electrons, so six electrons are transferred in total [2]. [2]

(c) Two bracketed X³⁺ ions and three bracketed Y^2- ions [2]; each Y^2- has eight outer-shell electrons [1], with two transferred electrons shown on each [2]; all charges and the 2:3 ratio are correct [1]. [6]

Exam-style question 7

Stretch note: This question connects Atomic Structure II electron configurations with ionic formulae, electron origin and conductivity. Atom X has electron configuration 1s² 2s² 2p⁶ 3s². Atom Y has electron configuration 1s² 2s² 2p⁵.

(a) Identify X and Y. [2 marks]

(b) State the ions formed by X and Y. [2 marks]

(d) Draw a complete electron-origin diagram for the compound. [5 marks]

(e) Explain why the compound does not conduct as a solid but does conduct when molten. [2 marks]

Total: 12 marks

Show answer & marking points

(a) X has 12 electrons, so X is magnesium, Mg [1]. Y has 9 electrons, so Y is fluorine, F [1]. [2]

(b) Mg loses two electrons to form Mg²⁺ [1]; F gains one electron to form F^- [1]. [2]

(d) One bracketed Mg²⁺ and two bracketed F^- ions [2]; each fluoride has eight outer-shell electrons [1], including one electron shown as transferred from Mg [1]; final electron totals and charges are correct [1]. [5]

(e) In solid MgF₂, ions are fixed and cannot carry charge [1]. When molten, the ions are mobile and carry charge [1]. [2]

✓ DONE WHENYour ionic diagram shows electron origin, correct electron totals, brackets and charges.

Giant ionic lattices and their properties

Learning goalExplain melting, solubility and conductivity using ions and their mobility.

🔓 CRACK IT

An ionic solid is a giant repeating arrangement of positive and negative ions. Strong electrostatic attractions act throughout the lattice, so a lot of energy is needed to separate the ions on melting.

For conductivity, ask whether charged particles can move. In the solid, the ions are held in fixed positions. When molten, the ions can move and carry charge.

Many ionic compounds dissolve in water, although not all do. If an ionic compound forms an aqueous solution, its mobile ions can carry charge. Do not explain ionic conductivity using moving electrons.

Higher ion charges and smaller ionic radii generally strengthen the electrostatic attraction. For the same charges, smaller ions let the centres of opposite ions get closer together. Stronger attractions need more energy to overcome, so the melting point tends to be higher. Compare similar lattice arrangements and change one factor at a time.

Giant ionic lattice: A regular repeating arrangement of oppositely charged ions strongly attracted in all directions.

Property	Explanation
High melting / boiling points	Many strong electrostatic attractions throughout the lattice require much energy to overcome.
Solid does not conduct	Ions are fixed; no mobile charge carriers.
Molten or dissolved sample conducts	Ions are mobile and carry charge.
Solubility	Often soluble in water, but not always: new solvent-ion attractions must compensate for attractions disrupted.

EXAM TRAPMolten ionic compounds conduct through mobile ions, not delocalised electrons. A charged ion does not automatically make every ionic compound soluble.

Worked example

Comparing ion charges: MgO and NaF

MgO and NaF have similar ion sizes and comparable lattice arrangements. Why does MgO have the higher melting point?

Worked answer

Both have giant ionic lattices. MgO contains Mg²⁺ and O²⁻; NaF contains Na⁺ and F⁻.
The ion sizes are similar, so focus on charge: the ions in MgO have larger charges.
The electrostatic attractions between opposite ions are stronger in MgO.
More energy is needed to overcome these attractions on melting.

Higher charges → stronger electrostatic attraction → more energy → higher melting point.

Worked example

Comparing ion sizes: NaCl and KCl

NaCl and KCl have the same ion charges and comparable lattice arrangements. Na⁺ is smaller than K⁺. Why does NaCl have the higher melting point?

Worked answer

Both contain a 1+ metal ion and the same Cl⁻ ion, so charge does not explain the difference.
The smaller Na⁺ ion allows the centres of opposite ions to be closer together.
The electrostatic attraction between the ions is stronger in NaCl.
More energy is needed to overcome these attractions on melting.

Smaller ions, with charge unchanged → stronger electrostatic attraction → higher melting point.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Quick check 8

Which has the higher melting point, CaO or KCl? Explain your answer. [3]

Total: 3 marks

Show answer & marking points

CaO has the higher melting point. Ca²⁺ and O²⁻ have larger charges than K⁺ and Cl⁻ [1], giving stronger electrostatic attractions between oppositely charged ions [1]. More energy is required to overcome these attractions when CaO melts [1].

Quick check 9

NaF and NaCl have comparable lattice arrangements. Explain why NaF has a higher melting point than NaCl. [3]

Total: 3 marks

Show answer & marking points

The smaller F⁻ ion allows the centres of opposite ions to be closer together [1]. With the same ion charges, electrostatic attractions are stronger in NaF [1], so more energy is needed to overcome them on melting [1].

Exam-style question 10

Sodium chloride forms a giant ionic lattice.

(a) Explain why sodium chloride has a high melting point. [2 marks]

(b) Explain why solid sodium chloride does not conduct electricity but molten sodium chloride does. [3 marks]

(d) State whether all ionic compounds are soluble in water. [1 mark]

Total: 7 marks

Show answer & marking points

(a) NaCl has many strong electrostatic attractions between oppositely charged ions throughout a giant lattice [1]; much energy is required to overcome them [1]. [2]

(b) In solid NaCl, ions are fixed in position, so there are no mobile charge carriers [1]. When molten, the ions are mobile [1] and carry charge through the liquid [1]. [3]

(d) Ionic compounds are often, but not always, soluble in water; solubility depends on whether new water-ion attractions compensate for separating the ions in the lattice [1]. [1]

Exam-style question 11

Each statement contains a chemistry error. Identify the error and write a corrected statement.

(a) The ionic bond is the transfer of electrons from a metal atom to a non-metal atom. [2 marks]

(b) Molten sodium chloride conducts because its electrons become free to move. [2 marks]

(d) All ionic compounds dissolve in water simply because their ions are charged. [2 marks]

Total: 8 marks

Show answer & marking points

(a) Transfer forms the ions [1]; ionic bonding is the electrostatic attraction between oppositely charged ions [1]. [2]

(b) The charge carriers are mobile Na⁺ and Cl^- ions [1], not free electrons [1]. [2]

(c) Melting iodine overcomes intermolecular forces between I₂ molecules [1]; the I-I covalent bonds remain intact [1]. [2]

(d) Ionic compounds are often but not always water-soluble [1]; solubility depends on the balance between attractions broken and new solvent-ion attractions formed [1]. [2]

✓ DONE WHENYou explain every property using the correct particles, attraction and mobility or energy.

Covalent, dative & bond strength

8 practice questions

Covalent bonds and dot-and-cross diagrams

Learning goalGive the full OCR definition and draw complete single and multiple bonds.

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In a covalent bond, two atoms share a pair of electrons. The shared electrons are attracted to both positive nuclei, holding the atoms together.

A single bond contains one shared pair, a double bond two and a triple bond three. In a dot-and-cross diagram, put every shared pair in the overlap between the atoms.

A lone pair is not shared in a bond. Show these electrons outside the overlap, and use dots and crosses consistently to show their atom of origin. Count all the outer-shell electrons before finishing.

Covalent bond: The strong electrostatic attraction between a shared pair of electrons and the nuclei of the bonded atoms.

Bond	Shared pairs	Line model
Single	1	H–H
Double	2	O=O
Triple	3	N≡N

EXAM TRAP“A shared pair of electrons” alone is incomplete as the full definition: include strong electrostatic attraction to both nuclei.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 12

(a) State the precise OCR A definition of a covalent bond. [3 marks]

(b) State the number of shared electron pairs in H₂, O₂ and N₂. [2 marks]

Total: 5 marks

Show answer & marking points

(a) A covalent bond is the strong electrostatic attraction [1] between a shared pair of electrons [1] and the nuclei of the bonded atoms [1]. [3]

(b) H₂ has one shared pair; O₂ has two; N₂ has three [2]. [2]

Exam-style question 13

Draw complete dot-and-cross diagrams for HCl, O₂ and N₂. Show all shared pairs and all lone pairs. [7 marks]

Total: 7 marks

Show answer & marking points

HCl: one shared dot-and-cross pair and three lone pairs on Cl [2]. O₂: two shared pairs and two lone pairs on each O [2]. N₂: three shared pairs and one lone pair on each N [3]. [7]

Exam-style question 14

Carbon dioxide, CO₂, contains two carbon-oxygen double bonds.

(a) State the total number of shared electron pairs in one CO₂ molecule. [1 mark]

(b) Draw a complete dot-and-cross diagram for CO₂. Show electron origin and all lone pairs. [6 marks]

Total: 7 marks

Show answer & marking points

(a) Two C=O double bonds contain four shared electron pairs in total [1]. [1]

(b) Carbon is central with two shared pairs to each oxygen [2]; electron origin is shown consistently in all four shared pairs [2]; each oxygen has two lone pairs and carbon has no lone pairs [2]. [6]

✓ DONE WHENYou can define a covalent bond precisely and account for every outer-shell electron.

Dative bonds: donor, acceptor and arrow

Learning goalShow how one atom supplies both electrons in a shared pair.

🔓 CRACK IT

A covalent bond contains a shared pair of electrons. Usually, each atom supplies one electron. In a dative covalent bond, one atom supplies both.

The atom supplying the pair is the donor. The species receiving it is the acceptor. Draw the arrow from the donor's lone pair to the acceptor.

For NH₃ reacting with H⁺, nitrogen supplies its lone pair to the hydrogen ion. Once NH₄⁺ has formed, all four N-H bonds are equivalent. The origin of the electrons does not make one of those bonds different afterwards.

Dative covalent bond: A shared pair of electrons in which both electrons were supplied by the same atom.

Role	NH₃ + H⁺ → NH₄⁺
Donor	Nitrogen in NH₃ supplies its lone pair.
Acceptor	H⁺ accepts the pair.
After formation	All four N–H bonds are equivalent.

EXAM TRAPStart the arrow at the donated electron pair and point towards the acceptor. The charge and brackets belong around the whole product ion.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 15

(a) State what is meant by a dative covalent bond. [2 marks]

(b) For NH₃ + H⁺ → NH₄⁺, name the electron-pair donor and acceptor and state the arrow direction. [2 marks]

(c) Draw the ammonium ion, showing the origin of the electron pair in the newly formed bond, square brackets and overall charge. [2 marks]

Total: 6 marks

Show answer & marking points

(a) A dative covalent bond is a shared pair of electrons [1] in which both electrons were supplied by the same atom [1]. [2]

(b) NH₃ is the electron-pair donor and H⁺ is the acceptor [1]; the arrow begins at the N lone pair and points towards H⁺ [1]. [2]

(c) NH₄⁺ is enclosed in square brackets with an overall + charge [1]; the new N-H shared pair contains two electrons originating from N [1]. [2]

Exam-style question 16

A water molecule reacts with H⁺ to form H₃O⁺.

(a) Identify the electron-pair donor and acceptor and state the direction of the dative-bond arrow. [3 marks]

(b) Draw the H₃O⁺ ion, showing the origin of the electron pair in the dative bond, brackets and charge. [3 marks]

Total: 6 marks

Show answer & marking points

(a) H₂O is the donor [1]; H⁺ is the acceptor [1]; the arrow starts at an oxygen lone pair and points to H⁺ [1]. [3]

(b) H₃O⁺ is bracketed and has an overall + charge [1]; the newly formed O-H shared pair contains two electrons from O [1]; the remaining outer electrons are complete and consistently shown [1]. [3]

Exam-style question 17

A student writes calcium nitrate as CaNO₃ and ammonium sulfate as NH₄SO₄.

Correct both formulae and explain the charge balance in each compound. [4]

The same student draws the following diagram for NH₃ + H⁺ → NH₄⁺.

(b) Identify two errors in the diagram. [2]

Total: 8 marks

Show answer & marking points

(a) Calcium nitrate is Ca(NO₃)₂: one Ca²⁺ requires two NO₃^- ions [2]. Ammonium sulfate is (NH₄)₂SO₄: two NH₄⁺ ions balance one SO₄^2- ion [2]. [4]

(b) The arrow is reversed; it must begin at the nitrogen lone pair and point to H⁺ [1]. The product must be enclosed in square brackets with an overall + charge [1]. [2]

✓ DONE WHENYou identify donor and acceptor before drawing the correct arrow and complete product ion.

Average bond enthalpy and bond strength

Learning goalCompare covalent bond strengths from supplied average bond enthalpies.

🔓 CRACK IT

Bond strength tells you how hard it is to separate the bonded atoms. A larger average bond enthalpy means more energy is needed to break the bond, so the bond is stronger.

Use the values supplied to make a comparison. Do not confuse the strength of a covalent bond inside a molecule with the attractions between separate molecules.

At this stage, use the comparison. Formal definitions and bond-enthalpy calculations are developed in Energetics.

Larger average bond enthalpy → stronger covalent bond → more energy needed to break it

Worked example

Compare two bonds

Bond A has average bond enthalpy 347 kJ mol^-1; bond B has 612 kJ mol^-1.

State which bond is stronger and explain your choice using the bond enthalpies.

Worked answer

612 is the larger value, so more energy is required to break a mole of B bonds.

Bond B is stronger.

EXAM TRAPDo not reverse the relationship: a large positive bond-breaking energy means a strong bond.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Quick check 18

(a) State the precise definition of metallic bonding. [2 marks]

(b) Bond X has an average bond enthalpy of 436 kJ mol^-1 and bond Y has 598 kJ mol^-1. Identify the stronger bond and justify your choice. [1 mark]

Total: 3 marks

Show answer & marking points

(a) Metallic bonding is the strong electrostatic attraction [1] between positive metal ions and delocalised electrons [1]. [2]

(b) Bond Y is stronger because it has the larger average bond enthalpy, so more energy is needed to break it [1]. [1]

Exam-style question 19

Average bond enthalpy data are shown: C–C 347 kJ mol^-1; H–H 436 kJ mol^-1; C=C 612 kJ mol^-1; N≡N 945 kJ mol^-1.

(a) Identify the strongest bond and explain your choice. [2 marks]

(b) Use the data to compare the strengths of a C–C bond and a C=C bond. [2 marks]

Total: 4 marks

Show answer & marking points

(a) N≡N is strongest because 945 kJ mol^-1 is the largest average bond enthalpy [1], so it requires the most energy to break [1]. [2]

(b) C=C is stronger than C-C [1] because 612 kJ mol^-1 is greater than 347 kJ mol^-1 [1]. [2]

✓ DONE WHENYou identify the stronger bond and justify it using the energy needed to break it.

Metallic bonding

1 practice questions

Learning goalExplain metallic properties using cations and delocalised electrons.

🔓 CRACK IT

A metal contains a giant lattice of positive ions surrounded by delocalised electrons. 'Delocalised' means the electrons are not restricted to one atom or one bond.

The attraction between the positive ions and these electrons holds the metal together. Strong attractions often give metals high melting and boiling points.

The delocalised electrons can move and carry charge in both solid and liquid metals. The positive ions do not travel through the solid to conduct electricity.

Metals do not normally dissolve as intact metallic lattices in molecular solvents. A metal reacting with a substance is a different process.

Metallic bonding: Strong electrostatic attraction between positive metal ions and delocalised electrons.

EXAM TRAPName the mobile charge carriers as delocalised electrons. The positive ions are fixed in a solid metal.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 20

Magnesium has a giant metallic lattice.

(a) Describe the particles and attraction in metallic magnesium. [2 marks]

(b) Explain why magnesium conducts electricity as both a solid and a liquid. [2 marks]

Total: 6 marks

Show answer & marking points

(a) Magnesium contains positive metal ions and delocalised electrons [1] with strong electrostatic attraction between them [1]. [2]

(b) Delocalised electrons are mobile in both solid and liquid magnesium [1] and carry electrical charge [1]. [2]

(c) Strong electrostatic attractions act throughout the giant lattice [1], so much energy is needed to overcome them [1]. [2]

✓ DONE WHENYou explain metallic conductivity in both states and link melting to electrostatic attraction.

Structures & properties

6 practice questions

Diamond and graphite

Learning goalLink covalent networks, layers and delocalised electrons to properties.

🔓 CRACK IT

Diamond and graphite both contain giant covalent networks, but each carbon atom bonds differently. That difference explains their different properties.

In diamond, each carbon forms four bonds in a rigid three-dimensional network. There are no mobile electrons, so it does not conduct.

In graphite, each carbon forms three bonds within a layer. One electron per carbon is delocalised and can carry charge. Weak van der Waals attractions between layers allow the layers to slide.

Both structures have high melting points because many strong covalent bonds must be broken. Graphite is not made of separate molecules, even if an older diagram calls the between-layer attractions 'intermolecular'. Silicon and graphene are also giant covalent networks; their fuller comparisons belong to Periodicity.

Giant covalent structure: A continuous network of atoms joined by many strong covalent bonds.

Feature	Diamond	Graphite
Bonding	Four covalent bonds per carbon	Three covalent bonds per carbon
Structure	Rigid three-dimensional network	Flat layers with weak attractions between them
Melting	Very high: many strong bonds must be broken	Very high: many strong bonds within layers must be broken
Conductivity	No mobile charge carriers	Delocalised electrons move along layers
Mechanical property	Hard, rigid network	Layers can slide; relatively soft

EXAM TRAPGraphite does not conduct because it has mobile ions. Sliding layers do not break the strong covalent bonds within each layer.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 21

Diamond and graphite are giant covalent forms of carbon.

(a) Explain why diamond has a very high melting point. [2 marks]

(b) Explain why diamond does not conduct electricity. [2 marks]

(d) Explain why the layers in graphite can slide over one another. [2 marks]

Total: 8 marks

Show answer & marking points

(a) Diamond is a giant three-dimensional covalent network [1]; many strong covalent bonds must be broken, requiring much energy [1]. [2]

(b) All four outer electrons of each carbon are localised in covalent bonds [1], so there are no mobile charged particles [1]. [2]

(c) Each carbon in graphite forms three covalent bonds, leaving delocalised electrons [1] that move along the layers and carry charge [1]. [2]

(d) Only weak attractions act between graphite layers [1], so little energy is needed for layers to slide [1]. [2]

Exam-style question 22

Graphite is used as an electrode. Diamond is used on cutting tools.

(a) Explain why graphite is suitable for use as an electrode. [3 marks]

(b) Explain why diamond is suitable for use on a cutting tool. [3 marks]

Total: 8 marks

Show answer & marking points

(a) Graphite contains delocalised electrons [1] that are mobile along the layers [1] and carry electrical charge through the electrode [1]. [3]

(b) Diamond is a rigid three-dimensional network [1] in which each carbon forms four strong covalent bonds [1], making it very hard and resistant to deformation [1]. [3]

✓ DONE WHENYou explain the similarities and differences using structure, bonds and mobile electrons.

Simple molecular lattices and state changes

Learning goalDistinguish bonds within molecules from attractions between them.

🔓 CRACK IT

A simple molecular substance contains separate molecules. Each molecule has strong covalent bonds inside it, while neighbouring molecules attract through intermolecular forces.

Melting and boiling overcome attractions between molecules; the molecules usually stay intact. This is why a molecular substance can have a low boiling point despite having strong covalent bonds.

Pure molecular substances normally have no mobile ions or delocalised electrons, so they do not conduct. To explain solubility, compare the attractions disrupted with the new attractions formed with the solvent.

Simple molecular solid: Discrete covalently bonded molecules held together by intermolecular attractions.

EXAM TRAPDo not say iodine melts because its I–I covalent bonds break. Detailed intermolecular force types follow in Bonding II.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 23

Iodine is a simple molecular substance made from I₂ molecules.

(a) Describe the particles and attractions in solid iodine. [2 marks]

(b) Explain why iodine has a much lower melting point than diamond. [2 marks]

(d) Explain why iodine does not conduct electricity in the solid or liquid state. [1 mark]

Total: 6 marks

Show answer & marking points

(a) Solid iodine contains neutral I₂ molecules [1] attracted to one another by intermolecular forces [1]. [2]

(b) Melting iodine overcomes only weak intermolecular (van der Waals) forces between I₂ molecules [1]. Much more energy is required to break the strong C–C covalent bonds in diamond [1]. [2]

(d) Neutral molecules provide no mobile ions or delocalised electrons [1]. [1]

✓ DONE WHENYou name which attractions change during melting and explain the lack of charge carriers.

Identify structures from physical properties

Learning goalCombine conductivity and melting-point evidence to identify a structure.

🔓 CRACK IT

Start a structure question by identifying the particles: ions, atoms or molecules. Then name the attraction between them. This stops you using the right explanation for the wrong structure.

Choose the structure type from the evidence: giant ionic, metallic, giant covalent or simple molecular.
For melting or boiling, name the attractions that must be overcome and connect their strength to the energy needed.
For conductivity, identify the charged particles and say whether they can move.
Use all the evidence together. A solid conductor could be a metal or graphite, so one property alone may not identify it.

Structure	Particles	Main attraction	Conductivity	Melting / boiling
Giant ionic	Positive and negative ions	Electrostatic attraction between ions	Molten / aqueous only	High
Metallic	Cations and delocalised electrons	Ions attracted to delocalised electrons	Solid and liquid	Often high
Giant covalent	Atoms	Strong covalent bonds throughout a network	Usually none; graphite is an exception	Very high
Simple molecular	Molecules	Intermolecular forces	None in pure substance	Usually lower

Worked example

An unknown solid

High melting point; does not conduct as a solid; conducts when molten.

Deduce the likely structure type of the solid and explain how its properties support your answer.

Worked answer

Charged particles exist, but they become mobile only after the solid lattice breaks down.

The evidence supports a giant ionic structure.

EXAM TRAPUse all the data. No conductivity alone cannot distinguish a molecular solid from diamond.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 24

The properties of four substances are shown.

Substance	Melting point / °C	Conducts as solid?	Conducts as liquid?
A	801	No	Yes
B	650	Yes	Yes
C	3550	No	No
D	114	No	No

Identify the most likely structure type for A–D. [4]
For each substance, use the conductivity evidence together with its melting point to justify the identification. [4]

Total: 8 marks

Show answer & marking points

(a) A ionic [1]; B metallic [1]; C giant covalent [1]; D simple molecular [1]. [4]

(b) A conducts only when its ions become mobile in the liquid [1]. B conducts in both states because delocalised electrons remain mobile [1]. C has no mobile charged particles [1]. D contains neutral molecules and no mobile charge carriers [1]. [4]

Conductivity alone cannot distinguish C from D: neither conducts. Their very different melting points provide the additional evidence.

Exam-style question 25

Four unfamiliar substances have the following properties.

Substance	Melting point / °C	Solid conductivity	Liquid conductivity
P	80	None	None
Q	1414	None	None
R	770	None	Good
S	1085	Good	Good

Identify the most likely structure type for P–S. [4]
Explain the change in conductivity when R melts. [3]
Explain why Q has a much higher melting point than P. [2]

Total: 9 marks

Show answer & marking points

(a) P simple molecular [1]; Q giant covalent [1]; R ionic [1]; S metallic [1]. [4]

(b) In solid R, ions are fixed and cannot carry charge [1]. On melting, ions become mobile [1] and carry charge through the liquid [1]. [3]

(c) Q requires many strong covalent bonds in a giant network to be broken [1], whereas P requires only intermolecular forces between molecules to be overcome [1]. [2]

✓ DONE WHENYou justify an identification using more than just a structure label.

Build complete property explanations

Learning goalConnect structure, particles, attractions, energy or mobility and the property.

🔓 CRACK IT

Build an explanation as a linked chain. Each sentence should explain why the next part follows.

Name the structure and the particles present.
Identify the attraction between those particles.
For a melting-point question, link attraction strength to the energy needed to separate the particles.
For a conductivity question, name the charge carrier and say whether it can move.

Use specific terms such as ions, delocalised electrons or intermolecular forces. A statement that a substance 'has strong bonds' does not identify the attraction.

Structure → particles → attraction → energy or mobility → property

Worked example

Sodium chloride versus iodine

Explain why sodium chloride has a much higher melting point than iodine. Refer to their structures and the attractions overcome on melting.

Worked answer

NaCl is giant ionic; iodine is simple molecular.
Strong electrostatic attractions act between ions throughout the NaCl lattice.
I₂ molecules are attracted by weaker intermolecular forces.
Much more energy is needed to overcome the ionic attractions.

NaCl has the higher melting point.

EXAM TRAPAvoid “ionic molecules”, “covalent bonds melt” and “all ionic compounds dissolve”. Use the named particles and their attraction.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 26

A student compares barium fluoride, BaF₂, potassium metal and iodine.

State the structure type of each substance. [3]
Explain why solid potassium conducts electricity but solid barium fluoride and solid iodine do not. [3]
Explain why barium fluoride has a much higher melting point than iodine. [2]

Total: 8 marks

Show answer & marking points

Barium fluoride: giant ionic [1]; potassium: giant metallic [1]; iodine: simple molecular [1].
Solid BaF₂ has fixed ions and no mobile charge carriers [1]. Potassium has mobile delocalised electrons that carry charge [1]. Neutral I₂ molecules provide no mobile ions or delocalised electrons [1].
BaF₂ has strong electrostatic attractions between Ba²⁺ and F⁻ ions throughout its lattice [1]. These require much more energy to overcome than the weak intermolecular forces between I₂ molecules [1].

✓ DONE WHENYou can compare structures without skipping a link in the explanation.

Shapes & bond angles

4 practice questions

Predict shape by counting electron regions

Learning goalCount bonding regions and lone pairs before naming the molecular shape.

🔓 CRACK IT

The shape comes from electron regions repelling one another around the central atom. They arrange as far apart as possible. Count bonding regions and lone pairs before naming the shape.

Find the central atom's outer-shell electron number: for example, 5 for N or 6 for O.
For a suitable species with single bonds to H or halogens, add one electron for each attached atom.
Add electrons for a negative charge and subtract them for a positive charge.
Divide by two to get the total pairs. Subtract the bonding pairs to find the lone pairs.
Choose the parent electron arrangement. Then name the molecular shape using the positions of the atoms.

The shortcut has limits. With multiple bonds or a supplied electron diagram, count regions directly: a single, double or triple bond is one bonding region. Lone pairs count as regions too.

Bonding pair: A pair of electrons shared between two covalently bonded atoms.
Lone pair: A pair of outer-shell electrons not used in a covalent bond.
Electron region: One direction of electron density: a single, double or triple bond counts as one region.
Molecular shape: The arrangement of atoms around the central atom.
Bond angle: The angle between two bonds meeting at the central atom.

Total pairs = (central valence electrons + attached H or halogens + negative charge − positive charge) ÷ 2

Total regions	Parent arrangement	Ideal angles
2	Linear	180°
3	Trigonal planar	120°
4	Tetrahedral	109.5°
5	Trigonal bipyramidal	90°, 120° and 180°
6	Octahedral	90° and 180°

Worked example

Count the pairs in ammonia

Predict the shape of NH₃.

State its bond angle and explain your prediction in terms of bonding pairs and lone pairs.

Worked answer

Nitrogen contributes five outer electrons: (5 + 3)/2 = 4 pairs.
Three pairs form N–H bonds, leaving one lone pair.
Four regions have a tetrahedral parent arrangement; the lone pair compresses the bonds.

Pyramidal, with a 107° H–N–H angle.

EXAM TRAPThe shortcut uses the number of outer-shell electrons, not modern group number 15 or 16. For multiple bonds, inspect the structure and count each multiple bond as one region.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 27

(a) State the electron-pair repulsion principle used to predict molecular shape. [2 marks]

(b) Place these repulsions in decreasing order of strength: bond pair-bond pair, lone pair-bond pair, lone pair-lone pair. [3 marks]

Total: 6 marks

Show answer & marking points

(a) Electron pairs around a central atom repel [1] and adopt positions of maximum separation / minimum repulsion [1]. [2]

(b) lone pair-lone pair [1] > lone pair-bond pair [1] > bond pair-bond pair [1]. [3]

✓ DONE WHENYou distinguish total electron regions, bonding regions and lone pairs and minimise their repulsion.

Linear, trigonal planar and tetrahedral shapes

Learning goalRecognise two, three and four bonding regions without lone pairs.

🔓 CRACK IT

When every electron region is a bond, the arrangement of the regions gives the molecular shape directly: two gives linear, three gives trigonal planar and four gives tetrahedral.

The bond angles are 180°, 120° and 109.5° respectively. Count a multiple bond as one region when deciding the shape.

A tetrahedral diagram needs depth. Use ordinary lines for bonds in the paper, a solid wedge towards you and a hashed wedge away. Keep equivalent bond lengths consistent and finish each bond at its atom label.

Regions	Shape	Angle	Examples
2	Linear	180°	CO₂, BeCl₂
3	Trigonal planar	120°	BF₃
4	Tetrahedral	109.5°	CH₄, NH₄⁺

Worked example

Why is CO2 linear?

Each C=O bond in CO₂ is a double bond. Predict the molecular shape and O–C–O bond angle, and explain them using electron-pair repulsion.

Worked answer

Each double bond occupies one direction of electron density. The two regions reach maximum separation on opposite sides of carbon.

Linear, 180°.

EXAM TRAPA tetrahedral molecule is three-dimensional, not four flat bonds at right angles.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 28

Complete the table by giving the molecular shape and bond angle(s). Names and angles only; diagrams are not required. [12]

Species	Bonding pairs	Lone pairs	Shape	Bond angle(s)
BeCl₂	2	0	…	…
BF₃	3	0	…	…
CH₄	4	0	…	…
NH₃	3	1	…	…
H₂O	2	2	…	…
SF₆	6	0	…	…

Total: 12 marks

Show answer & marking points

BeCl₂: linear [1], 180° [1]. BF₃: trigonal planar [1], 120° [1].

CH₄: tetrahedral [1], 109.5° [1]. NH₃: pyramidal [1], 107° [1].

H₂O: non-linear [1], 104.5° [1]. SF₆: octahedral [1], 90° between adjacent bonds [1]. [12]

✓ DONE WHENYou can name, draw and give the angles for the three bonding-only arrangements.

Lone pairs compress bond angles

Learning goalExplain pyramidal and non-linear shapes using stronger lone-pair repulsion.

🔓 CRACK IT

A lone pair takes up more space near the central atom than a bonding pair. It is attracted to one nucleus rather than shared between two nuclei. It repels nearby bonding regions more strongly and pushes them closer together.

CH₄, NH₃ and H₂O each have four electron regions around the central atom. Replacing bonds with lone pairs changes the positions of the atoms and therefore the molecular shape.

CH₄: four bonds, no lone pairs; tetrahedral, 109.5°.
NH₃: three bonds, one lone pair; pyramidal, about 107°.
H₂O: two bonds, two lone pairs; bent/non-linear, about 104.5°.

Subtracting about 2.5° for each lone pair is a memory aid for this familiar comparison. It is not a universal formula for every molecule or parent arrangement.

Terminology: pyramidal and trigonal pyramidal are two names for this required shape. NH₃ is the common example, with a bond angle of about 107°.

Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair

Molecule	Regions	Shape	Angle
NH₃	3 bonding + 1 lone	Pyramidal (trigonal pyramidal)	107°
H₂O	2 bonding + 2 lone	Non-linear (bent)	104.5°

Worked example

Why is water’s angle smaller?

State the H–O–H bond angle in water and the H–N–H bond angle in ammonia. Explain why the angle in water is smaller.

Worked answer

Water has two lone pairs; ammonia has one.
Lone pairs repel more strongly than bonding pairs.
The O–H bonds are pushed closer together.

104.5° in H₂O compared with 107° in NH₃.

EXAM TRAPThe familiar 109.5° → 107° → 104.5° sequence is not a universal “subtract 2.5° per lone pair” rule. Use known examples or supplied data for other species.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 29

(a) State the numbers of bonding pairs and lone pairs around the central atom in NH₃ and H₂O. [2 marks]

(b) For each molecule, state the shape name and draw a clear 3-D diagram. Also state its bond angle. [4 marks]

Total: 8 marks

Show answer & marking points

(a) NH₃: three bonding pairs and one lone pair [1]. H₂O: two bonding pairs and two lone pairs [1]. [2]

(b) NH₃: correctly named and drawn pyramidal shape [1], 107° [1]. H₂O: correctly named and drawn non-linear shape [1], 104.5° [1]. [4]

✓ DONE WHENYou name the molecular shape from atom positions and explain the angle using lone-pair repulsion.

Molecular ions and up to six regions

Learning goalInclude charge in the electron count and recognise five- and six-region arrangements.

🔓 CRACK IT

Treat a molecular ion with the same method as a neutral molecule. Include the charge when counting electrons, then separate the bonding regions from the lone pairs.

Four regions give NH₄⁺ a tetrahedral shape because all four are bonds. In H₃O⁺, one region is a lone pair, so the atom arrangement is pyramidal.

The named shapes include linear, non-linear, trigonal planar, pyramidal (also called trigonal pyramidal), tetrahedral and octahedral.

Trigonal bipyramidal, as in PCl₅, is a different shape and is not in that named list. You must still apply electron-pair repulsion to molecules and ions with up to six electron pairs around the central atom.

Species	Bonding + lone regions	Shape	Angles
NH₄⁺	4 + 0	Tetrahedral	109.5°
H₃O⁺	3 + 1	Pyramidal	No fixed recall angle stated here
SF₆	6 + 0	Octahedral	90° adjacent, 180° opposite
PCl₅	5 + 0	Trigonal bipyramidal	120° equatorial; 90° axial-equatorial; 180° axial-axial

Worked example

Account for the charge in NH4+

Predict the shape of the ammonium ion.

State its bond angle and explain your prediction in terms of electron pairs.

Worked answer

Electron total = 5 + 4 − 1 = 8; divide by two to get four pairs.
All four pairs are bonding pairs, so they repel equally.
Draw the ion in square brackets with an overall + charge.

Tetrahedral, 109.5°.

EXAM TRAPDo not give H₃O⁺ an angle merely by copying NH₃. An ion’s charge belongs around the whole species.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 30

(a) PCl₅ has five bonding pairs and no lone pairs around P. SF₆ has six bonding pairs and no lone pairs around S. State the name of the shape and the bond angle(s) for each molecule. [6 marks]

(b) Draw a clear 3-D diagram of each molecule to match the shape names in part (a). Use wedge-and-dash bonds where needed and keep equivalent bonds similar in length. [2 marks]

Total: 8 marks

Show answer & marking points

(a) PCl₅: trigonal bipyramidal [1], with 120° between equatorial bonds [1], and 90° between axial and equatorial bonds plus 180° between the two axial bonds [1]. SF₆: octahedral [1], with 90° between adjacent bonds [1] and 180° between opposite bonds [1]. [6]

(b) One chemically correct 3-D drawing for each molecule, with the five or six bonds arranged correctly [1 each]. [2]

✓ DONE WHENYou handle charge, brackets and up to six electron regions in a three-dimensional drawing.

Electronegativity & polarity

5 practice questions

Electronegativity and polar bonds

Learning goalUse Pauling values to explain unequal sharing and partial charges.

🔓 CRACK IT

Electronegativity tells you how strongly an atom attracts the shared pair in a covalent bond. Unequal attraction makes the bond polar: the more electronegative atom is δ− and its partner is δ+.

Electronegativity generally increases across a period and up a group. Fluorine is the most electronegative element.

Calculate a difference by subtracting the smaller value from the larger one. Always use the numbers supplied in that question: some practice questions use H = 2.1, so they need not give the same numerical difference as a different notes example.

A partial charge is not an ionic charge. The atoms are still sharing the bonding pair.

Electronegativity: The ability of an atom to attract the bonding electrons in a covalent bond.
Polar bond: A covalent bond with an unequal distribution of bonding electron density because the atoms have different electronegativities.
Permanent dipole: A permanent separation of partial charge, represented by δ+ and δ−.

Atom	H	C	N	O	F	Cl
Approximate Pauling value for these notes	2.2	2.5	3.0	3.5	4.0	3.0

Worked example

Compare C–H and O–H

The Pauling electronegativities are C = 2.5, H = 2.2 and O = 3.5. Calculate the electronegativity difference for C–H and O–H. State which bond is more polar and identify the δ+ and δ− atoms in that bond.

Worked answer

ΔEN(C–H) = 2.5 − 2.2 = 0.3.
ΔEN(O–H) = 3.5 − 2.2 = 1.3.
The larger difference gives more unequal sharing.

O–H is more polar; O is δ− and H is δ+.

EXAM TRAPδ is a partial charge, not a full ionic charge. Bond character changes continuously; no single numerical electronegativity threshold is an absolute law.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 31

(a) Define electronegativity using precise OCR A wording. [2 marks]

(b) State which element is the most electronegative and describe the general direction in which electronegativity increases in the periodic table. [2 marks]

(c) Pauling electronegativities are H 2.1, C 2.5, O 3.5 and F 4.0. Calculate the electronegativity difference for C-H and O-H, then identify the more polar bond. [3 marks]

Total: 7 marks

Show answer & marking points

(a) Electronegativity is the ability of an atom to attract the bonding electrons [1] in a covalent bond [1]. [2]

(b) Fluorine is the most electronegative element [1]; electronegativity generally increases towards fluorine, across a period and up a group [1]. [2]

(c) C-H: |2.5 - 2.1| = 0.4 [1]. O-H: |3.5 - 2.1| = 1.4 [1]. O-H is more polar because it has the larger electronegativity difference [1]. [3]

Exam-style question 32

(a) Explain what is meant by a polar covalent bond and a permanent dipole. [2 marks]

(b) Use δ+ and δ- to show the permanent dipole in H-Cl. Add a bond-dipole arrow pointing in the correct direction. [2 marks]

Total: 6 marks

Show answer & marking points

(a) A polar covalent bond has an unequal distribution of bonding electron density caused by different electronegativities [1], producing permanent partial charges / a permanent dipole [1]. [2]

(b) H is δ+ and Cl is δ- [1]; the dipole arrow points from H towards Cl, with the crossed tail at the H end [1]. [2]

(c) Bond polarity concerns one bond [1]. Molecular polarity is the vector result of all bond dipoles in the 3-D shape; they may reinforce or cancel [1]. [2]

Exam-style question 33

Selected Pauling electronegativities are Mg 1.2, H 2.1, C 2.5, Cl 3.0, O 3.5 and F 4.0.

(a) Calculate the electronegativity difference for Mg-Cl, C-Cl and O-H. [3 marks]

(b) Rank these three bonds from least polar to most polar. [2 marks]

(d) State which atom is δ- in a C=O bond. [1 mark]

Total: 8 marks

Show answer & marking points

(a) Mg-Cl: 3.0 - 1.2 = 1.8 [1]; C-Cl: 3.0 - 2.5 = 0.5 [1]; O-H: 3.5 - 2.1 = 1.4 [1]. [3]

(b) C-Cl < O-H < Mg-Cl [2]. [2]

(c) Mg-Cl has a much larger electronegativity difference, indicating much greater ionic character [1]; C-Cl has a smaller difference and the electrons are shared covalently [1]. [2]

(d) O is δ- because it is more electronegative than C [1]. [1]

✓ DONE WHENYou define electronegativity precisely, calculate differences and orient partial charges and dipole arrows correctly.

Bond polarity and molecular polarity

Learning goalCombine bond dipoles with the three-dimensional shape to decide overall polarity.

🔓 CRACK IT

A molecule can contain polar bonds and still be non-polar overall. The answer depends on whether the bond dipoles cancel in its three-dimensional shape.

Identify which bonds are polar and mark their dipole directions.
Work out the molecular shape, including the effect of lone pairs.
Check whether the bond dipoles balance in all directions.
If they cancel completely, the molecule has no permanent dipole. If they do not, the molecule is polar.

CO₂ is linear, so its equal C=O dipoles cancel. H₂O is bent, so its O-H dipoles do not cancel. Draw the shape before deciding.

Polar molecule: A molecule with an overall dipole because its bond dipoles do not cancel in its molecular shape.

Molecule	Shape / dipole arrangement	Overall polarity
HCl	One bond dipole	Polar
CO₂	Linear, equal opposite dipoles	Non-polar
BF₃	Symmetrical trigonal planar	Non-polar
CH₄	Symmetrical tetrahedral	Non-polar
NH₃	Pyramidal; dipoles do not cancel	Polar
H₂O	Bent; dipoles do not cancel	Polar
SF₆	Symmetrical octahedral	Non-polar

Worked example

H2O versus CO2

H₂O and CO₂ both contain polar covalent bonds. State which molecule is polar overall and explain the difference using molecular shape and bond dipoles.

Worked answer

CO₂ is linear, so the equal C=O dipoles point in opposite directions and cancel. H₂O is bent, so its O–H dipoles do not cancel.

CO₂ has no overall dipole; H₂O is polar.

EXAM TRAP“Symmetrical” alone is not a complete explanation. State which equal dipoles cancel and how they are arranged.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 34

(a) For each molecule, state whether it is polar or non-polar. [5]

Molecule	Polar / non-polar
CO₂	…
BF₃	…
H₂O	…
NH₃	…
CCl₄	…

(b) Choose one non-polar molecule from part (a) that contains polar bonds. State its shape, then explain why it has no overall dipole. A drawing is not required. [3]

Total: 8 marks

Show answer & marking points

(a) CO₂ non-polar [1]; BF₃ non-polar [1]; H₂O polar [1]; NH₃ polar [1]; CCl₄ non-polar [1]. [5]

(b) Example CO₂: each C=O bond is polar [1], but the molecule is linear and symmetrical [1], so equal dipoles act in opposite directions and cancel [1]. [3]

Exam-style question 35

Carbon dioxide, O=C=O, and hydrogen cyanide, H-C≡N, are both linear. Pauling electronegativities are H 2.1, C 2.5, N 3.0 and O 3.5.

(a) State the shape name and bond angle around carbon in each molecule. Names and angles only; diagrams are not required in part (a). [4 marks]

(b) Add bond-dipole arrows to both molecules. Include partial charges at the terminal atoms. [2 marks]

Total: 8 marks

Show answer & marking points

(a) CO₂ is linear, 180° [2]. HCN is linear around carbon, 180° [2]. [4]

(b) Correct terminal partial charges and dipole directions towards O in CO₂ and towards C then N across H-C≡N [1 each molecule]. [2]

(c) CO₂ has equal opposite C=O dipoles that cancel [1]. HCN is unsymmetrical, so its directed bond dipoles do not cancel and it has an overall dipole [1]. [2]

✓ DONE WHENYou decide overall polarity from bond dipoles and geometry rather than the presence of one polar bond.

Intermolecular forces

4 practice questions

Bonds within molecules and forces between them

Learning goalDistinguish intramolecular bonds from intermolecular attractions.

🔓 CRACK IT

Keep the location of an attraction clear. Covalent bonds hold atoms together within a molecule. Intermolecular forces attract one molecule to another.

Start with induced dipole-dipole attractions, which act between all atoms and molecules. Polar molecules also have permanent dipole-dipole attractions. Suitable molecules can form hydrogen bonds as well.

List all the forces present before comparing substances. Naming hydrogen bonding does not mean the other applicable forces disappear.

London dispersion is another name for induced dipole-dipole interactions. Van der Waals is also used as a broader term for induced and permanent dipole-dipole interactions; the specific name makes your meaning clearer.

Intramolecular bond: A bond within a particle, such as an O–H covalent bond within water.
Intermolecular force: An electrostatic attraction between separate molecules.
Volatility: The tendency to vaporise; higher volatility usually accompanies a lower boiling point.

Interaction	Where it acts	Cause
Induced dipole-dipole (London dispersion)	All atoms and molecules	Instantaneous dipoles induce neighbouring dipoles.
Permanent dipole-dipole	In addition, between polar molecules	Opposite permanent partial charges attract.
Hydrogen bonding	H directly bonded to N/O/F and a neighbouring N/O/F lone pair	A lone pair attracts a strongly δ+ hydrogen on another molecule.

Worked example

What changes when iodine boils?

When iodine boils, identify the particles that separate and the attractions overcome. State whether the I–I covalent bonds break.

Worked answer

Iodine contains discrete I₂ molecules.
Induced dipole-dipole attractions act between the molecules.
Boiling overcomes those intermolecular attractions.

The molecules separate; I–I covalent bonds remain intact.

EXAM TRAPDo not replace a named intermolecular force with “strong bonds”, which could be mistaken for covalent bonds.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 36

(a) Define an induced dipole-dipole interaction. You may also use the term London dispersion force. [2 marks]

(b) Define a permanent dipole-dipole interaction. [2 marks]

Total: 6 marks

Show answer & marking points

(a) A fluctuation in electron density creates an instantaneous dipole [1], which induces an opposite dipole in a neighbouring particle and causes attraction [1]. [2]

(b) Opposite partial charges on neighbouring polar molecules attract [2]. [2]

(c) Hydrogen bonding is an intermolecular attraction involving H bonded to N, O or F [1] and a lone pair on N, O or F in another molecule [1]. [2]

Exam-style question 37

(a) State the strongest intermolecular force present between molecules or atoms of each substance. Use the most specific OCR A term. [5]

Substance	Strongest force
Ne	…
HCl	…
NH₃	…
CH₃OH	…
CO₂	…

(b) Explain why HCl molecules also experience induced dipole-dipole interactions. [2]

Total: 7 marks

Show answer & marking points

(a) Ne: induced dipole-dipole [1]; HCl: permanent dipole-dipole [1]; NH₃: hydrogen bonding [1]; CH₃OH: hydrogen bonding [1]; CO₂: induced dipole-dipole [1]. [5]

(b) Every electron cloud fluctuates [1], so HCl also forms instantaneous dipoles that induce dipoles in neighbouring molecules [1]. [2]

✓ DONE WHENYou name the interaction correctly and keep it separate from bonds inside each molecule.

Induced dipole-dipole interactions

Learning goalExplain how electron fluctuations create attraction and affect boiling points.

🔓 CRACK IT

Electrons are always moving. For an instant, they can be distributed unevenly, giving one side of a particle a small negative charge and the other a small positive charge.

This instantaneous dipole distorts the electron cloud of a neighbouring particle.
That distortion creates an induced dipole in the neighbour.
Opposite partial charges attract. New dipoles continually form as the electrons move.

A larger electron cloud is usually more easily distorted, or more polarisable. This tends to give stronger van der Waals attractions and a higher boiling point.

For isomers with the same electron count, more branching usually reduces surface contact between molecules. The overall attractions are then weaker, so less energy is needed to separate them.

Induced dipole-dipole interaction: Attraction between an instantaneous dipole in one particle and the dipole it induces in a neighbour.

Factor	Effect
More electrons / larger cloud	Usually more polarisable, so larger temporary dipoles and stronger attractions.
Greater molecular contact	More electron-cloud contact can strengthen the overall attraction.

Worked example

I2 compared with Br2

Why does I₂ have the higher boiling point?

Explain your answer in terms of intermolecular attractions and state which substance is more volatile.

Worked answer

I₂ has more electrons and a larger, more polarisable cloud. Stronger induced dipole-dipole attractions require more energy to overcome.

I₂ has the higher boiling point and lower volatility.

EXAM TRAPAll molecules have these interactions, including polar molecules. Explain the electron cloud rather than saying only “the molecule is heavier”.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 38

The boiling points of three halogens are shown.

Substance	Boiling point / °C
Cl₂	−34.0
Br₂	58.8
I₂	184.3

State the strongest intermolecular force between molecules of each halogen. [1]
Explain the boiling-point trend from Cl₂ to I₂. [4]
Predict whether F₂ boils above or below −34.0 °C. Justify your answer. [2]

Total: 7 marks

Show answer & marking points

(a) Induced dipole-dipole / London dispersion forces [1]. [1]

(b) From Cl₂ to I₂ the molecules contain more electrons and have larger, more polarisable electron clouds [1]; instantaneous dipoles are larger [1], so induced dipole-dipole attractions are stronger [1] and more energy is needed to separate the molecules [1]. [4]

(c) F₂ should boil below -34.0 °C [1] because it has fewer electrons / a less polarisable cloud and weaker induced dipole-dipole attractions than Cl₂ [1]. [2]

✓ DONE WHENYou connect electron movement, polarisability, attraction and energy to the boiling-point trend.

Permanent dipole-dipole interactions

Learning goalExplain attractions between polar molecules while retaining their induced interactions.

🔓 CRACK IT

A polar molecule has a permanent uneven distribution of charge. The δ+ end of one molecule attracts the δ− end of a neighbouring molecule.

These permanent dipole-dipole attractions act as well as induced dipole-dipole attractions. For otherwise comparable molecules, the extra attractions can increase the energy needed to separate them.

HCl has an overall dipole because it has one polar bond. SO₂ is bent, so its bond dipoles do not cancel. Use the shape as well as bond polarity when deciding whether this force is present.

Do not rank boiling points from polarity alone. A large non-polar molecule can have stronger overall attractions than a small polar molecule.

Permanent dipole-dipole interaction: Electrostatic attraction between the permanent δ+ end of one polar molecule and the permanent δ− end of a neighbour.

Molecule	Interactions
Cl₂ or linear CO₂	Induced dipole-dipole
HCl or bent SO₂	Induced plus permanent dipole-dipole

Worked example

HCl compared with F2

HCl and F₂ have similar numbers of electrons. Explain why HCl has the higher boiling point.

Worked answer

Their electron counts provide a useful starting comparison for induced interactions.
HCl is polar; F₂ is non-polar.
HCl also has permanent dipole-dipole attractions.

More energy is required to separate HCl molecules.

EXAM TRAPPermanent dipoles do not automatically outweigh a much larger electron cloud. Compare the whole set of interactions.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 39

(a) Draw two HCl molecules to show a permanent dipole-dipole attraction. Use solid lines for H-Cl bonds, a dotted line for the intermolecular attraction and correct partial charges. [4 marks]

(b) Name the dotted attraction. [1 mark]

Total: 7 marks

Show answer & marking points

(a) Two solid H-Cl covalent bonds [1]; H δ+ and Cl δ- on both molecules [1]; a dotted line between Cl δ- and H δ+ [1]; molecules aligned so opposite partial charges attract [1]. [4]

(b) Permanent dipole-dipole attraction [1]. [1]

(c) No electron pair is shared between the molecules [1]; the dotted line is an electrostatic intermolecular attraction and is much weaker than the intramolecular H-Cl covalent bond [1]. [2]

✓ DONE WHENYou draw correctly aligned partial charges and distinguish the dotted attraction from a covalent bond.

Hydrogen bonding & properties

10 practice questions

Hydrogen bonding and correct diagrams

Learning goalIdentify a suitable δ+ H atom and neighbouring lone pair and draw a clear hydrogen bond.

🔓 CRACK IT

Hydrogen bonding needs a particular arrangement. Find a δ+ hydrogen directly bonded to N, O or F, then a lone pair on N, O or F in a neighbouring molecule.

Find a δ+ H atom directly bonded to N, O or F in one molecule.
Find a lone pair on N, O or F in a neighbouring molecule.
Draw a dotted attraction from that lone pair to the δ+ H. The covalent bond from H to its own atom stays in place.
Add the appropriate partial charges and keep the hydrogen-bond arrangement clear.

Water, ammonia and HF can hydrogen-bond between their own molecules. An oxygen lone pair can attract a δ+ H atom in water, even when the oxygen-containing molecule has no O-H bond and cannot hydrogen-bond with itself. H bonded only to carbon does not meet the hydrogen-bonding condition here.

Hydrogen bond: An intermolecular attraction between a lone pair on N, O or F in one molecule and a δ+ H atom covalently bonded to N, O or F in another molecule.

Check	Requirement
Hydrogen atom	H directly bonded to N, O or F.
Lone pair	An available lone pair on N, O or F.
Drawing	Solid covalent bonds; dotted intermolecular line from the neighbouring N/O/F lone pair to δ+ H; correct δ signs.

Worked example

Water compared with hydrogen sulfide

Why does H₂O have a much higher boiling point than H₂S?

Worked answer

H₂O has O–H bonds and oxygen lone pairs, so its molecules hydrogen-bond.
H₂S does not form the same hydrogen-bonding network.
More energy is needed to overcome the attractions between water molecules.

H₂O boils higher and is less volatile.

EXAM TRAPKeep the N/O/F lone pair, dotted line, H and its covalently bonded N/O/F approximately collinear. Do not draw a solid bond between molecules.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 40

(a) Draw two water molecules and show one hydrogen bond. Include δ+ and δ-, a lone pair, solid covalent bonds and a dotted intermolecular attraction. [5 marks]

(b) Identify the two atoms linked by the hydrogen bond in your diagram, and explain why the dotted line is not an O-H covalent bond. [2 marks]

Total: 7 marks

Show answer & marking points

(a) Two correct bent H₂O molecules with solid O-H bonds [1]; O δ- and H δ+ [1]; a lone pair on O [1]; dotted H···O attraction [1]; line joins δ+ H to a lone pair / δ- O of the other molecule [1]. [5]

(b) The hydrogen bond joins the δ+ H of one water molecule to the lone pair on O of another [1]. The dotted line is an intermolecular electrostatic attraction, not a shared electron pair / covalent bond [1]. [2]

Exam-style question 41

The boiling points of four hydrogen halides are shown.

Substance	Boiling point / °C
HF	19.5
HCl	−85
HBr	−67
HI	−35

Explain why HF has a much higher boiling point than HCl. [4]
Explain the increase in boiling point from HCl to HI. [4]

Total: 8 marks

Show answer & marking points

(a) F is highly electronegative, so HF has a strongly polar H-F bond [1]. The H δ+ of one molecule is attracted to a lone pair on F of another [1], forming hydrogen bonds [1]. More energy is required to overcome these stronger attractions [1]. [4]

(b) HCl to HI have progressively more electrons / larger electron clouds [1], which are more easily polarised [1]. Induced dipoles and London forces therefore become stronger [1], so more energy is needed and boiling point rises [1]. [4]

Exam-style question 42

Ethanol, CH₃CH₂OH, and methoxymethane, CH₃OCH₃, have the same molecular formula and relative molecular mass. Their boiling points are 78.4 °C and -24.8 °C respectively.

(a) State the strongest intermolecular force in each pure liquid. [2 marks]

(b) Explain the large difference in boiling point. [4 marks]

Total: 8 marks

Show answer & marking points

(a) Ethanol: hydrogen bonding [1]. Methoxymethane: permanent dipole-dipole attraction [1]. [2]

(b) Ethanol contains an O-H bond and can form hydrogen bonds between its molecules [1]. Methoxymethane has no H directly bonded to O, N or F and cannot hydrogen-bond with itself [1]. Hydrogen bonds are stronger than its permanent dipole-dipole attractions [1], so more energy is required to separate ethanol molecules [1]. [4]

Exam-style question 43

Hydrazine has the structural formula H₂N-NH₂. Each nitrogen atom has three bonded pairs and one lone pair. Hydrazine boils at 114 °C; ethane, C₂H₆, boils at -89 °C. Their relative molecular masses are similar.

(a) State the shape name around each nitrogen atom and state how its bond angle compares with 109.5°. A drawing is not required in part (a). [3 marks]

(b) Draw one hydrogen bond between two hydrazine molecules. Show the N-H bond, the neighbouring nitrogen lone pair and partial charges. [2 marks]

Total: 7 marks

Show answer & marking points

(a) Each N has four electron pairs, one of which is lone, so the shape around each N is pyramidal [2]. Lone-pair repulsion compresses the angle below 109.5° [1]. [3]

(b) Dotted N-H···N attraction from δ+ H in an N-H bond [1] to a lone pair on the neighbouring δ− N [1]. [2]

(c) Hydrazine forms hydrogen bonds whereas ethane has only induced dipole-dipole attractions [1]; more energy is needed to separate hydrazine molecules [1]. [2]

Exam-style question 44

A student’s proposed hydrogen-bond diagram for two ethanol molecules is shown.

Identify three errors in the diagram. [3]
Redraw the interaction correctly. [3]

Total: 6 marks

Show answer & marking points

(a) O atoms should be δ-, not δ+ [1]; H in O-H should be δ+, not δ- [1]; the intermolecular attraction must be dotted / broken rather than drawn as a solid covalent-bond line [1]. [3]

(b) Correct solid covalent bonds [1], δ+ H···O δ- dotted attraction [1] and a lone pair on oxygen in the neighbouring molecule [1]. [3]

✓ DONE WHENYou can identify every part of a hydrogen bond and explain its effect on properties.

Water’s boiling point and the density of ice

Learning goalUse hydrogen bonding to explain water’s anomalous properties.

🔓 CRACK IT

Hydrogen bonds help explain two unusual properties of water. They make water's boiling point higher than the simple trend for comparable hydrides suggests, and they give ice its open structure.

In ice, hydrogen bonds hold water molecules further apart than they are on average in liquid water. The same mass occupies a larger volume, so ice is less dense and floats.

On melting, some hydrogen bonds are overcome and the open arrangement partly collapses. The O-H covalent bonds inside each water molecule remain intact.

Use the same link for boiling-point comparisons: stronger overall attractions between molecules mean more energy is needed to separate them.

Anomaly	Explanation
Relatively high melting and boiling points	Many hydrogen bonds require a substantial energy input to overcome enough attractions for molecules to move or separate.
Ice is less dense than liquid water	An ordered open lattice holds water molecules farther apart. Part of the network collapses on melting.

Worked example

Why does ice float?

Explain why solid ice is less dense than liquid water and therefore floats on water.

Worked answer

Hydrogen bonds hold water molecules in an open, ordered lattice.
For the same mass, ice occupies a larger volume.
Density = mass ÷ volume, so ice is less dense.

Ice floats on the denser liquid.

EXAM TRAPO–H covalent bonds remain intact during melting and boiling. The diagram shows selected links, not a complete crystal unit cell.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 45

(a) Draw part of the open hydrogen-bonded lattice in ice. Distinguish O-H covalent bonds from hydrogen bonds. [3 marks]

(b) Explain why ice is less dense than liquid water. [3 marks]

Total: 8 marks

Show answer & marking points

(a) Water molecules are linked in an open arrangement by dotted hydrogen bonds [1], with solid O-H covalent bonds clearly distinguished [1] and the dotted attraction joining H δ+ to O δ- / a lone pair [1]. [3]

(b) In ice, hydrogen bonds hold water molecules in a fixed open lattice [1] with relatively large spaces [1]. The same mass therefore occupies a larger volume than in liquid water, so ice has lower density [1]. [3]

✓ DONE WHENYou connect hydrogen bonding both to the energy needed for state changes and to the spacing of molecules.

Molecular lattices, solubility and volatility

Learning goalExplain state changes, conductivity, solubility and volatility using molecular attractions.

🔓 CRACK IT

For a molecular substance, separate the strong covalent bonds inside the molecules from the attractions between them. Melting and boiling normally overcome the between-molecule attractions.

Stronger overall attractions tend to increase boiling point and reduce volatility. Neutral molecules normally provide no mobile ions or delocalised electrons, so the pure molecular substance does not conduct.

For solubility, identify the attractions disrupted and the new ones formed with the solvent. A molecule can accept a hydrogen bond from water without being able to hydrogen-bond to itself. The examples below show how to name the relevant interactions.

Simple molecular lattice: Discrete covalently bonded molecules held together in a solid by intermolecular forces.

Property	Explanation
Melting / boiling	Intermolecular forces are overcome; molecules usually stay intact.
Pure electrical conductivity	No mobile ions or delocalised electrons.
Solubility	Depends on new solute-solvent attractions compared with those disrupted.
Volatility	Weaker attractions usually mean a lower boiling point and higher volatility.

Worked example

Iodine versus diamond

Explain why iodine has a much lower melting point than diamond. Refer to their structures and the attractions that must be overcome.

Worked answer

Iodine is simple molecular, so melting overcomes attractions between I₂ molecules. Diamond is a giant covalent network whose strong C–C bonds would need to be broken.

Much more energy is required for diamond.

Worked example

Iodine in cyclohexane

Why is iodine much more soluble in cyclohexane than in water?

Worked answer

I₂ and cyclohexane are non-polar. Favourable induced dipole-dipole attractions can form between them, whereas iodine does not replace water’s hydrogen-bond attractions as effectively.

Iodine is more soluble in cyclohexane.

Worked example

Methanol in water

Why does methanol mix completely with water?

Worked answer

Methanol has an O–H group and oxygen lone pairs. Methanol and water form favourable hydrogen bonds with one another.

Methanol is miscible with water.

EXAM TRAP“Like dissolves like” is a prediction, not an absolute law. A molecular solute conducts only if mobile ions are present, for example after reaction or ionisation.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 46

Consider iodine, I₂, methanol, CH₃OH, and hexane, C₆H₁₄.

(a) Predict whether iodine is more soluble in water or in hexane. Explain your answer using intermolecular forces. [3 marks]

(b) Predict whether methanol is more soluble in water or in hexane. Explain your answer. [3 marks]

Total: 7 marks

Show answer & marking points

(a) I₂ is more soluble in hexane [1]. Both are non-polar and interact mainly through induced dipole-dipole forces [1]; comparable new attractions can replace those broken on mixing [1]. [3]

(b) Methanol is more soluble in water [1]. Both are polar and methanol can form hydrogen bonds with water [1], giving favourable solute-solvent attractions [1]. [3]

✓ DONE WHENYou justify each physical property with particles and attractions rather than a slogan.

Build complete intermolecular-force comparisons

Learning goalCompare electron clouds, permanent dipoles and hydrogen bonding before discussing energy.

🔓 CRACK IT

To compare boiling points, explain what has to be overcome when the particles separate. Stronger overall intermolecular attractions need more energy, so boiling point tends to be higher.

Identify the molecules and all the forces between them.
Compare electron count, polarisability, shape and any extra permanent-dipole or hydrogen-bonding attractions.
Decide which substance has stronger overall attractions.
Link that to more energy needed and the higher boiling point.

A more volatile substance evaporates more readily. Stronger attractions generally reduce volatility at a given temperature. Melting points also depend on how well particles pack in the crystal, so avoid using one simple force ranking for every comparison.

Particles → named attraction → relative strength → energy → property

Weak explanation	Improve it
“NH₃ has stronger bonds”	Name hydrogen bonds between molecules, not N–H covalent bonds.
“I₂ is bigger”	Explain electron count, polarisability, induced dipoles, attraction and energy.
“CO₂ bonds are non-polar”	The C=O bonds are polar; equal opposite dipoles cancel.

Worked example

Ammonia compared with phosphine

Why does NH₃ have a higher boiling point than PH₃?

Worked answer

NH₃ contains N–H bonds and a nitrogen lone pair, so its molecules hydrogen-bond. PH₃ does not. More energy is needed to separate NH₃ molecules.

NH₃ has a higher boiling point and lower volatility.

EXAM TRAPMolecular polarity is one factor. Larger electron clouds can make a non-polar substance boil higher than a smaller polar one.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 47

Each carbon atom in the following chloromethanes is approximately tetrahedral.

Molecule	Boiling point / °C
CH₃Cl	−24
CH₂Cl₂	40
CHCl₃	61
CCl₄	77

Explain why every C–Cl bond is polar. [2]
Determine which molecules are polar. For the non-polar molecule, name its shape and explain dipole cancellation. A drawing is not required. [4]
Explain why CCl₄ can have the highest boiling point despite being non-polar. [2]

Total: 8 marks

Show answer & marking points

(a) Cl is more electronegative than C [1], so electron density is drawn towards Cl and each C-Cl bond has C δ+ and Cl δ- [1]. [2]

(b) CH₃Cl, CH₂Cl₂ and CHCl₃ are polar [3]. CCl₄ is tetrahedral and symmetrical, so its four equal C-Cl dipoles cancel and it is non-polar [1]. [4]

(c) CCl₄ has the largest electron cloud / most electrons [1], so it has the strongest induced dipole-dipole forces and needs the most energy to separate its molecules [1]. [2]

Exam-style question 48

Three substances have similar relative molecular masses but very different boiling points.

Substance	Formula	Mᵣ	Boiling point / °C
Ethane	C₂H₆	30	−89
Methanal	CH₂O	30	−19
Methanol	CH₃OH	32	65

State the strongest intermolecular force in each substance. [3]
Explain the boiling-point order, distinguishing all three types of intermolecular force. [4]

Total: 7 marks

Show answer & marking points

(a) Ethane: induced dipole-dipole [1]. Methanal: permanent dipole-dipole [1]. Methanol: hydrogen bonding [1]. [3]

(b) Similar Mr values reduce the difference in London forces [1]. Methanal has a permanent C=O dipole, giving stronger attraction than in non-polar ethane [1]. Methanol molecules form hydrogen bonds [1], the strongest interaction in the set, so methanol requires the most energy to boil [1]. [4]

✓ DONE WHENEvery comparison names the interactions before linking their strength to energy and the observed property.

Combine shape, polarity and physical properties

Learning goalApply the full chain to unfamiliar molecules.

🔓 CRACK IT

A longer question may connect shape, polarity and boiling point. Work through those ideas in that order so one answer supports the next.

Count the central-atom electron regions, including lone pairs.
Draw and name the three-dimensional molecular shape.
Use that shape to decide whether the bond dipoles cancel.
Identify all the applicable intermolecular forces, including induced dipoles.
Explain the property using energy needed to separate particles, or the mobility of charge carriers.

Keep the reasoning visible. A correct shape or a named force on its own is only part of the explanation.

Electron regions → shape → bond dipoles → overall polarity → intermolecular forces → property

Worked example

BF3 has polar B–F bonds

BF₃ contains polar B–F bonds. Predict its molecular shape, bond angle and overall polarity, explaining your reasoning.

Worked answer

Boron has three bonding regions and no lone pairs.
The regions arrange trigonal planar at 120°.
Three equal B–F dipoles are symmetrically arranged and cancel.

BF₃ is trigonal planar, 120° and non-polar overall.

EXAM TRAPDo not decide molecular polarity from the name of a bond, or boiling point from polarity alone.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 49

Molecules X, Y and Z are CH₃F, CF₄ and CH₃OH respectively. Their boiling points are -78 °C, -128 °C and 65 °C. All three have four electron regions around carbon.

(a) For both CH₃F and CF₄, state the shape name and bond angle around carbon, then draw a clear 3-D diagram. [3 marks]

(b) Explain why CH₃F is polar but CF₄ is non-polar. [3 marks]

(d) Explain the boiling-point order Y < X < Z. [2 marks]

Total: 10 marks

Show answer & marking points

(a) Both molecules are tetrahedral around carbon with a bond angle of approximately 109.5° [1]. Award one mark for each of two clear 3-D wedge-and-dash diagrams with four bonds and similar bond lengths [2]. [3]

(b) In CH₃F, the polar C-F bond produces a resultant dipole that is not cancelled by the three C-H bond dipoles [2]. In CF₄, four identical C-F dipoles are arranged symmetrically and cancel [1]. [3]

(c) X CH₃F: permanent dipole-dipole; Y CF₄: induced dipole-dipole; Z CH₃OH: hydrogen bonding. Award [2] for all three correct or [1] for any two. [2]

(d) CF₄ has only induced dipole forces; CH₃F also has permanent dipole attractions; CH₃OH forms hydrogen bonds, so progressively more energy is needed to separate the molecules [2]. [2]

✓ DONE WHENYou can link the whole chain without confusing covalent bonds with attractions between molecules.

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